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13. Logarithmic functions We saw in the previous chapter that e x was a one–to–one function. This means that it has an inverse function—this we will call lnÝ x Þ . It is also commonly referred to as logÝ x Þ or loge Ý x Þ (log to the base e of x). Thus the solution of the equation ex = 2 can be written as x= This is not necessarily very helpful, since we have to work out the value of ln 2 . However, most scientific calculators have a button for this purpose. More generally y = ex é x = lnÝ y Þ . (*) i.e., if fÝ x Þ = e x then f ?1 Ý x Þ = lnÝ x Þ . Since e x is always positive, we cannot solve the equation y = e x when y is either negative or zero—the range of e x is 0,K . From the discussion on page 3 of Chapter 10 we have the following table. Function x y=e , y = lnÝ x Þ , Domain Range Ý ?K,K Þ 0,K 0,K Ý ?K,K Þ We deduce from equation (*) that y = e lnÝy Þ ln e x for any positive number y = x for all x 5 R. These are the results of f f ?1 Ý y Þ = y and f ?1 fÝ x Þ = x. Note that because e x > 0 for all x, it follows that lnÝ y Þ is not defined for negative values of y. So ln ?2 , for example, cannot be evaluated without the use of complex numbers. Since e 0 = 1 we deduce that = ln 1 1 and, since e = e, lnÝ e Þ = 6 4 2 -6 -4 -2 0 2 x 4 6 -2 -4 -6 Graph of e x (dashed) and lnÝ x Þ (solid) More generally, if y = b x (where b is a postive base), then x = logb Ý x Þ —“log to base b of x”. We shall deal almost exclusively with natural logs (loge Ý x Þ , which is written lnÝ x Þ ). The natural log “ ln” could be changed to log of any base in most of the examples. The Laws of Logarithms Important rules for working with logarithms (these are valid for all logs, regardless of the base, so we write log rather than log10 ). Suppose that p and q are positive real numbers and that r is any real number. Then we have the following MA11001: Calculus Chapter 13 1 rules for manipulating logs. Theorem 13.1 a) logÝpqÞ = logÝpÞ + logÝqÞ (the log of a product is the sum of the logs) b) logÝ pq Þ = logÝpÞ ? logÝqÞ (the log of a quotient is the difference of the logs) c) log p r = r logÝpÞ (the log of a power is the exponent times the log). So, for example ln 5 + lnÝ x Þ = ln 5x ln 5 ? lnÝ x Þ = ln 5x 3ln 5 = ln 5 3 = ln 125 . Proof Let y = pq then, since e lnÝy Þ = y, e lnÝpq Þ = However, p = e lnÝp Þ and q = e lnÝq Þ and so (1) pq = where we have used e x 6 e t = e x+t . But this value of pq must coincide with that in equation (1) and so e lnÝpq Þ = The first rule then follows because e x is one–to–one: this means that when e x = e t then we must have x = t. The second rule can be deduced from the first: let x = pq , so we need to find lnÝ x Þ . However, p = xq and so, by the first rule lnÝ p Þ = which is rearranged to give rule b). For rule c) we again argue from two directions, as for rule a). First, with y = p r , r e lnÝp Þ= but, since p = e lnÝp Þ , (2) r pr = x = r where we have used e = e rx . But this value of p r must agree with that in equation (2) so, since e x is a one–to–one function, rule c) follows.n Note: there is no rule for expanding lnÝ p + q Þ . Exercise: Prove that lnÝ pqr Þ = lnÝ p Þ + lnÝ q Þ + lnÝ r Þ for any three positive real numbers p, q, r. Example 1. In Example 1 of Chapter 12 we saw that the number of doves at time t was 1000Ý1.1Þ t . Suppose we need to know the time at which the population reaches 3700. For this we need to solve the equation 3700 = 1000Ý1.1Þ t ì Ý1.1Þ t = 3.7 to find the value of t. The answer is to take logs of both sides: ln 1.1 t = ln 3.7 and to use rule c): from which we find t= Example 2. If the money in our account grows at 2% per year and we began with £5000 the amount we have at year n will be gÝ n Þ = 5000 1.02 n .(See Example 2 of Chapter 12). The money will double when 5000 1.02 n = i.e., 1.02 n = whose solution is obtained by taking logs of both sides: ln 1.02 n = ln 2 , so n= MA11001: Calculus Chapter 13 2 NB With a growth rate of 2%, it always takes the same number of years to double, regardless of starting amount. Example 3. From Example 3 of Chapter 12 the number of dogs in the population at year t was p t = 0.955 t 1500. The population will have halved when i.e., 0.955 t t = .955 1500 = 1. 2 Taking logs of both sides then gives t= NB With a decay rate of 4.5%, it always takes the same number of years to halve, regardless of starting amount. This time is known as the “ half–life” of the population. Example 4. On page 3 of Chapter 12 we saw that 10 u 2 3.3219... , where did this exponent come from? Solving 10 = 2 x by first taking logs of both sides gives ln 10 = x= and, with a calculator, ln 10 ln 2 = 3. 321928095... . Example 5. Solve 2 x 6 3 2?x = 5. We take logs of both sides to obtain, with the aid of rule a), We now apply rule c) twice on the left to give Collecting terms in x on the left and moving the remaining ones to the right gives x Example 6. Solve 2 x = 5 4 x . 3 We take logs of both sides and use rules a) and b) on the right, then use rule c) on both sides xln 2 = Collecting terms in x on the left and moving the remaining ones to the right gives Example 7. Solve 2 e 6 e x = e 2 6 2 x . We take logs of both sides and use rule a) then use rule c) where we have used lnÝ e Þ = 1.Collecting terms in x on the left and moving the remaining ones to the right gives Example 8. Solve 2a x = b 2x . Example 9. Solve the equation e x ? e ?x = 4. x ?x ¾ This time we let y = e then, since e = 1x , the equation becomes e MA11001: Calculus Chapter 13 3 from which we find y= Then, since e x = y, we have, on taking natural logs, x = lny. However, 2 ? 5 < 0 so there is no real value for x corresponding to y = 2 ? 5. Hence, the only solution is x = ln 2 + 5 . y = 1 ? 2x for y. 1?y Using the fact that e p = e q when p = q, we “take the exponential” of each side to give (we use the notation exp Ý x Þ instead of e x because the exponent is cumbersome) y exp ln = e 1?2x 1?y and, since exp lnÝ p Þ = p, this simplifies to Example 10. Solve ln Next multiply both sides by the denominator then collect all y terms on the left: Then y= 3 Example 11. Simplify lnÝ pq Þ ? 2lnÝ p Þ + ln qp , where p and q are positive quantities. By rule c) 2lnÝ p Þ = then, with rule b) lnÝ pq Þ ? then, with rule a) 3 lnÝ pq Þ ? 2lnÝ p Þ + ln qp Example 12. Simplify 2lnÝ xy Þ ? = 1 2 ln x 2 y + 4ln x where x and y are positive real numbers. 2lnÝ xy Þ = 1 2 ln x 2 y 4ln x = = and so 2lnÝ xy Þ ? 1 2 ln x 2 y + 4ln x = Example 13. Solve lnx 2 + lnx = ln27, x 5 0,K 2 lnx + lnx = and so the equation becomes and we have x= since 27 = 3 3 . Example 14. Solve the equation lnÝx ? 2Þ + lnÝ2x + 4Þ = ?2. MA11001: Calculus Chapter 13 4 First note that the maximal domain of lnÝ x Þ is x 5 interval 2,K . 0,K so any solution to this equation must lie in the Now combine the two terms on the left using the Rule: the sum of logs is the log of the product: ln Take the exponential of both sides exp ln so, since exp lnÝ y Þ = y, we obtain the quadratic equation which simplifies to x2 = There appear to be two solutions of this equation x=± However, as mentioned at the start of this solution, the natural logarithm is only defined for positive values of its argument. Both x ? 2 and 2x + 4 must be positive in the original equation, lnÝx ? 2Þ + lnÝ2x + 4Þ = ?2, so we have to discard the negative solution because Ý?2.01685 ? 2Þ < 0. The only valid solution is x = 4 + 12 e ?2 u 2.02. This is confirmed by drawing the graph of the functions y = lnÝx ? 2Þ + lnÝ2x + 4Þ and y = ?2 (shown below), since they cross just to the right of x = 2. 4 2 -2 2 x 4 6 0 -2 -4 -6 -8 -10 Example 15. Does the equation lnÝx ? 2Þ ? lnÝx + 1Þ = 5 have any real solutions? Again, note that any solution must lie in the interval 2,K . Next , combining the log terms using the rule: the difference of logs is the log of the quotient: = ln ¾ and taking exponentials of both sides (we use the notation exp Ý y Þ instead of e y because the exponent is cumbersome) exp = e5 . ln Using inverse relation exp lnÝ x Þ = x gives = e5 and so x= This would make x ? 2 = and x+1 = However, the natural log is only defined for positive values of its argument, so the conclusion is that there MA11001: Calculus Chapter 13 5 are no real solutions of the equation. The graphs of the functions y = lnÝx ? 2Þ ? lnÝx + 1Þ and y = 5 are shown below and do not cross. 4 2 -2 -1 1 2 x 3 4 5 0 -2 -4 -6 -8 -10 One should be wary when using Scientific Notebook to plot logarithmic functions. When asked to plot the function lnÝx ? 2Þ ? lnÝx + 1Þ it first simplifies using the rule of logs and actually plots the function ln x?2 x+1 which is shown below. It is evident that the equation ln x?2 does have a real solution. What is it? x+1 20 10 -4 0 -2 2 4 x -10 -20 The domain of fÝ x Þ = lnÝx ? 2Þ ? lnÝx + 1Þ is x 5 ?K,?1 W 2,K . 2,K while the domain of ln x?2 x+1 is The difficulties involved with logs of negative numbers are often avoided because the argument is enclosed in | 6 |. Example 16. Prove that ln|x | is an even function. With fÝ x Þ = ln|x | and bearing in mind the definition of the modulus (or absolute value) function |x | = x if x ³ 0 ?x if x < 0 which means that |x | is even: |?x | = |x |, we find fÝ ?x Þ = and so ln|x | is an even function. 4 2 -4 -2 0 2 x 4 -2 -4 Graph of ln|x |, Also, ln|x | ¸ ?K as x ¸ 0 so ln 0 x 5 R\ 0 . is undefined and the maximal domain of ln|x | is x 5 R\ 0 . Example 17. Find all real solutions of the equation ln 3x ? 2 ? ln x + 2 = 3. The presence of the absolute values means that the expression ln 3x ? 2 ? ln x + 2 is defined for all x 5 R\ 2/3,?2 : we have to exclude x = ?2 and x = 2/3 since both values lead to ln 0 which is undefined. Since MA11001: Calculus Chapter 13 6 ln 3x ? 2 ? ln x + 2 = ln we have so solve = e3 . There are two possibilities, = e3 which leads to u x= or = ?e 3 which leads to u x= That there are two solutions is confirmed by the graphs of y = ln 3x ? 2 ? ln x + 2 and y = 3. 10 8 6 4 2 -4 -3 -2 x -1 0 1 2 -2 -4 -6 -8 -10 The course text : Calculus by Anton , Section 4.2 has further exercises & examples. Common Logarithms Logarithms were invented by John Napier () of Edinburgh but it was Henry Briggs (1561–1630) who was responsible for publicising them and ensuring that they became widely used in scientific calculations. It was on his suggestion that Napier adjusted his definitions so that he used a base b = 10. That is, the inverse function of fÝ x Þ = 10 x is f ?1 Ý x Þ = log10 Ý x Þ y = 10 x é x = log10 Ý y Þ . These are now called common logs and were an extremely powerful tool for multiplication and division before electronic calculators became available. This is illustrated in the next example. Example Because of the laws of logarthims given earlier, common logarithms were used to speed up arithmetic calculations. For example, to multiply 1234.56 × 34.567 log10 Ý1234.56 × 34.567Þ = log10 Ý1234.56Þ + log10 Ý34.567Þ = 3.0915 + 1.5387 = 4.6302 4. 630 2 and so 1234.56 × 34.567 = 10 = 42677. 6, (which could be found by looking up a table of values of the inverse function). Logarithms convert a problem of multiplication into the easier tasks of looking up tables and adding: look up the logs of the two numbers in a book of tables, add the results together, then look up the inverse (called “anti-logs”). Similarly, division is reduced to subtraction. MA11001: Calculus Chapter 13 7 Common logs are especially important when measuring quantities that vary greatly in magnitude, since the common log of the numbers 10 ?6 , 10 ?3 ,1, 10 3 ,10 6 , for example, give the modestly sized numbers ?6, ?3, 0, 3 and 6. Practical examples of measuring units that use common logs are given on page 8. Example Moore’s Law. Below is a table of the speeds of the fastest supercomputer of the time—taken from SIAM News, Nov. 2001. The speed is given in “flops”: floating point operations per second which, loosely speaking, is the time it takes to perform a calculation such as “3x + 5” (i.e., one multiplication and one addition). Year Speed log10 speed 1950 7.0×10 2 2.84 1951 9.0×10 2 2.95 1960 6.0×10 4 4.78 1.0×10 6 6.00 6.0×10 6 6.78 6.0×10 6 6.78 1.0×10 8 8.00 1.0×10 9 9.00 2.0×10 9 9.30 1987 1.0×10 10 10.0 1991 1.3×10 11 11.1 1992 1.3×10 11 11.1 1998 1.3×10 12 12.1 2000 3.0×10 12 12.4 1965 1970 1971 1976 1982 1985 It can be seen that the speed varies enormously, from 700 flops to 3 × 10 11 flops. The common log of the speed however, varies only from 2.8 to 12.4. This shows up more dramatically when we plot this data. The top graph above plots speed against time and we can see that it has almost no information content. The speeds at earlier years are too small to register on the scale. What we then do is to plot log10 speed against time. This is shown in the bottom graph and, more clearly, on page 11 where the names of the computers are also given. The data points lie pretty well on a straight line and, with a little effort, one can find that the equation of the straight line through the data is approximately log10 speed where x = year ?1950. Since 10 log 10 speed u 2. 78 + 0.2x = speed, we find speedu 10 2. 78+0.2x u 602. × 10 0.2x . That is, speed u 600 × 10 year ?1950 /5 Þ To get a better grip on this equation, we observe that number of years it takes for the speed to double is t, where 10 t/5 = 2 i.e., t = 5log10 2 u 1. 51. This is Moore’s Law: Computer speed doubles roughly every 18months. MA11001: Calculus Chapter 13 8 Practical Applications Logarithms still have important uses now as the following example show. DECIBEL (dB) (excerpt from Audio Dictionary by Glenn While) Literally, one tenth of a bel. The bel is named after Alexander Graham Bell, and the number of bels is defined as the common logarithm of the ratio of two powers (as in electrical power). Signal Intensity Sound level in dB = 10log10 Reference Intensity A rule of thumb that is useful to remember is that 10 dB is a power ratio of 10. Any time a power is increased tenfold, it is increased by 10 dB; thus, a 200W power amplifier will put out 10 dB more electrical power than a 20W amplifier, and its sound power output will also increase by 10 dB. pH value in Chemistry: pH = ? log10 ßH +à = negative log of hydrogen ion concentration pH = 7 is neutral, pH < 7 is acid (Smaller number, more acidic) and pH > 7 is alkaline (Larger number, more alkaline.) Earthquakes and the Richter Scale Seismologists use a Magnitude scale to express the seismic energy released by each earthquake. Here are the typical effects of earthquakes in various magnitude ranges: Earthquake Severity (Richter scale) Effect < 3.5 Generally not felt, but recorded. 3.5 to 5.4 Often felt, but rarely causes damage. Under 6.0 At most slight damage to well-designed buildings. 6.1 to 6.9 Can be destructive in areas up to about 100 kilometers 7.0 to 7.9 Major earthquake. Can cause serious damage over larger areas. >8 Great earthquake. Can cause serious damage in areas several hundred kilometers across. If the amplitude of an earthquake is Amm (as measured by a Wood-Anderson seismograph), then Richter Magnitude = log10 ÝAÞ + ÝDistance Correction FactorÞ where the distance factor comes from the distance of the seismometer from the earthquake and is tabulated in Richter’s (1958) book “Elementary Seismology ”. MA11001: Calculus Chapter 13 9